Let $g(x)=x^{-10}$. $g'(1)=$
Solution: Let's first find the expression for $g'(x)$ and then evaluate it at $x=1$. The derivative of $g$ can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is a negative number.) $\begin{aligned} &\phantom{=}g'(x) \\\\ &=\dfrac{d}{dx}\left(x^{-10}\right) \\\\ &=-10x^{-10-1} \gray{\text{The power rule}} \\\\ &=-10x^{-11} \end{aligned}$ So we found that $g'(x)=-10x^{-11}$, which can also be written as $-\dfrac{10}{x^{11}}$. Now let's plug ${x=1}$ : $\begin{aligned} -\dfrac{10}{({1})^{11}}&=-\dfrac{10}{1} \\\\ &=-10 \end{aligned}$ In conclusion, $g'(1)=-10$.